Given that eccentricity i.e., $e=3 / 2$ and $( \pm a e, 0)=( \pm 2,0)$

$$\begin{array}{ll} \therefore & a e=2 \\ \Rightarrow & a \cdot \frac{3}{2}=2 \Rightarrow a=4 / 3 \\ \because & b^2=a^2\left(e^2-1\right) \\ \Rightarrow & b^2=\frac{16}{9}\left(\frac{9}{4}-1\right) \\ \Rightarrow & b^2=\frac{16}{4}\left(\frac{5}{4}\right)=+\frac{20}{9} \end{array}$$

So, the equation of hyperbola is

$$\begin{aligned} & \frac{x^2}{\frac{16}{9}}-\frac{y^2}{\frac{20}{9}}=1 \\ & \Rightarrow \quad \frac{x^2}{4}-\frac{y^2}{5}=\frac{4}{9} \end{aligned}$$

$$\begin{array}{r} & \text { Given lines are } & 2 x-3 y-5 =0 \quad \text{... (i)}\\ & \text { and } & 3 x-4 y-7 =0 \\ & \text { From Eqs. (i) and (ii), } & \frac{x}{21-20} =\frac{y}{-15+14}=\frac{1}{-8+9} \\ \Rightarrow & \frac{x}{1} =\frac{y}{-1}=\frac{1}{+1} \\ \Rightarrow & x = \pm 1, y=-1 \end{array}$$

Since the intersection point of these lines will be coordinates of the circle i.e., coordinates of the circle as $(1,-1)$.

Let the radius of the circle is $r$.

$$\begin{aligned} \text{Then,}\quad \pi r^2 & =154 \\ \Rightarrow \quad \frac{22}{7} \times r^2 & =154 \\ \Rightarrow\quad r^2 & =\frac{154 \times 7}{22} \\ \Rightarrow\quad r^2 & =\frac{14 \times 7}{2} \Rightarrow r^2=49 \end{aligned}$$

So, the equation of circle is

$$\begin{aligned} & (x-1)^2+(y+1)^2 =49 \\ \Rightarrow \quad & x^2-2 x+1+y^2+2 y+1 =49 \\ \Rightarrow \quad & x^2+y^2-2 x+2 y =47 \end{aligned}$$

Let the general equation of the circle is

$$x^2+y^2+2 g x+2 f y+c=0\quad \text{... (i)}$$

Since, this circle passes through the points $(2,3)$ and $(4,5)$.

$$\begin{array}{lrl} \therefore & 4+9+4 g+6 f+c & =0 \\ \Rightarrow & 4 g+6 f+c & =-13 \quad\text{... (ii)}\\ \text { and } & 16+25+8 g+10 f+c & =0 \\ \Rightarrow & 8 g+10 f+c & =-41\quad \text{... (iii)} \end{array}$$

Since, the centre of the circle $(-g,-f)$ lies on the straight line $y-4 x+3=0$

i.e., $\quad +4 g-f+3=0\quad \text{... (iv)}$

From Eq. (iv), $4 g=f-3$

On putting $4 \mathrm{~g}=f-3$ in Eq. (ii), we get

$$\begin{aligned} & f-3+6 f+c=-13 \\ \Rightarrow & 7 f+c=10\quad \text{.... (v)} \end{aligned}$$

From Eqs. (ii) and (iii),

$$\begin{gathered} 8 g+12 f+2 c=-26 \\ 8 g+10 f+c=-41 \\ \frac{-\quad-\quad-\qquad\quad+}{2 f+c=15}\quad \text{... (vi)} \end{gathered}$$

From Eqs. (ii) and (vi),

$$ \begin{gathered} 7 f+c=-10 \\ 2 f+c=15 \\ \frac{-\quad-\quad-}{5 f=-25} \end{gathered}$$

$$\begin{aligned} \therefore\quad f & =-5 \\ \text{Now,}\quad c & =10+15=25 \\ \text{From Eq. (iv),}\quad 4 g+5+3 & =0 \\ \Rightarrow\quad g & =-2 \end{aligned}$$

From Eq. (i), equation of the circle is $x^2+y^2-4 x-10 y+25=0$.

Given centre of the circle is $(3,-1)$.

$$\begin{aligned} \text{Now,}\quad& O P=\left|\frac{6+5+18}{\sqrt{4+25}}\right|=\frac{29}{\sqrt{29}}=\sqrt{29} \\ \text{In } \triangle O P B,\quad & O B^2=O P^2+P B^2 \quad {[\because A B=6 \Rightarrow P B=3]} \\ \Rightarrow \quad & O B^2=29+9 \Rightarrow O B^2=38 \end{aligned}$$

So, the radius of circle is $\sqrt{38}$,

$\therefore$ Equation of the circle with radius $r=\sqrt{38}$ and centre $(3,-1)$ is

$$\begin{array}{r} \Rightarrow \quad & (x-3)^2+(y+1)^2=38 \\ \Rightarrow \quad & x^2-6 x+9+y^2+2 y+1=38 \\ &x^2+y^2-6 x+2 y=28 \end{array}$$

Let the coordinates of centre of the required circle are $(h, k)$, then the centre of another circle is $(1,2)$.

Radius $=\sqrt{1+4+20}=5$

So, it is clear that $P$ is the mid-point of $C_1 C_2$.

$$\begin{array}{r} \therefore \quad & 5=\frac{1+h}{2} \Rightarrow h=9 \\ \text { and } \quad & 5=\frac{2+k}{2} \Rightarrow k=8 \end{array}$$

So, the equation of and required circle is

$$\begin{array}{r} & (x-9)^2+(y-8)^2 =25 \\ \Rightarrow \quad & x^2-18 x+81+y^2-16 y+64 =25 \\ \Rightarrow \quad & x^2+y^2-18 x-16 y+120 =0 \end{array}$$